Question

1. Givens s m (kg) F(N) 0.74 10.00 100 A. What is the magnitude of the force necessary to cause the impending motion of object? B. When F is exerted on the object, it causes an acceleration of 4.414 m/s2, what is / ? C. What is ufr when t-5s? D. If F is removed after t 5 s, how far will the object travel after this point?

Answer 1

To start the motion, applied force has to overcome maximum static friction possible which is given by f = mu_s mg therefore F = f = mu_s mg = (0.74)(10 kg)(9.8 m/s^2) = 72.52 n B) equation of motion is F - f = ma or, 100 N - mu+k (10 kg)(9.8 m/s^2) = (10kg)(4.414 m/s^2) or, mu_k = 0.57 c) V_fx = at = (4.414 m/s^2)(5s) = 22.07 m/s d) deceleration after F is removed is a = mu_k g = (0.57)(9.8 m/s^2) = 5.586 m/s^2 using v^2 = u^2 + 2as (u v_fx = 22.07 m/s for the portion of motion after t = 50) O = (22.07 m/s)^2 - 2(5.586 m/s^2)s or, s = 43.6 m
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